Painting the Walls

class Solution {
public:
    int calculateCost(vector<int>& costs, vector<int>& durations) {
        int totalTasks = costs.size();
        vector<int> currentCost = vector(totalTasks + 1, 0);
        vector<int> previousCost = vector(totalTasks + 1, (int) 1e9);
        previousCost[0] = 0;
            
        for (int task = totalTasks - 1; task >= 0; task--) {
            currentCost = vector(totalTasks + 1, 0);
            for (int remaining = 1; remaining <= totalTasks; remaining++) {
                int paintOption = costs[task] + previousCost[max(0, remaining - 1 - durations[task])];
                int skipOption = previousCost[remaining];
                currentCost[remaining] = min(paintOption, skipOption);
            }
                
            previousCost = currentCost;
        }
            
        return currentCost[totalTasks];
    }
};

In the provided C++ code, the goal is to determine the minimum cost to complete all given tasks of painting walls with dynamic programming. Each task has associated costs and durations that need adaptation based on efficiency and prior decisions. Below is the conceptual breakdown of the function calculateCost inside the Solution class:

• The function calculateCost accepts two vectors as parameters: costs and durations. Both vectors store integer values where each index represents a specific painting task. costs[i] represents the cost associated with task i, and durations[i] denotes the count of tasks that can be skipped after completing task i.

• Initial preparation:

  • Calculate totalTasks from the size of the costs vector.
  • Create currentCost and previousCost vectors with the size of totalTasks + 1. previousCost is initialized with high values (1e9) to simulate infinity, except for the first element which starts at 0. This setup helps in determining the minimum cost effectively by comparing newly calculated costs against practically infinite previous values.

• The solution utilizes a reverse iteration from totalTasks - 1 down to 0, recalculating costs for each task in consideration of dependencies introduced by durations.

  • For each task, the algorithm computes:
    • A paintOption cost that considers the cost of doing the current task combined with the minimum costs of tasks that can still be completed considering current task duration.
    • A skipOption that represents the cost of skipping the current task and sticking with the previously calculated costs.
    • The minimum of these options is then selected for the current stage in currentCost.

• Finally, the function returns the cost to complete all tasks obtained from the last iteration stored in currentCost[totalTasks].

This approach efficiently breaks down the problem of calculating the minimum cost with dependencies using dynamic programming, ensuring that all constraints and task sequences are considered. The vector handling and iterative comparisons ensure a streamlined calculation suitable for a range of possible task arrays.

class Solution {
    public int calculatePaintingCost(int[] costs, int[] durations) {
        int count = costs.length;
        int[] currentCost = new int[count + 1];
        int[] nextCost = new int[count + 1];
        Arrays.fill(nextCost, Integer.MAX_VALUE);
        nextCost[0] = 0;
    
        for (int j = count - 1; j >= 0; j--) {
            currentCost = new int[count + 1];
            for (int remains = 1; remains <= count; remains++) {
                int paint = costs[j] + nextCost[Math.max(0, remains - 1 - durations[j])];
                int skip = nextCost[remains];
                currentCost[remains] = Math.min(paint, skip);
            }
                
            nextCost = currentCost;
        }
            
        return currentCost[count];
    }
}

The "Painting the Walls" Java solution utilizes dynamic programming to calculate the minimum cost required to paint a set of walls while adhering to specified durations. Each wall has an associated cost and a duration that denotes the time it remains painted before requiring a repaint. The central theme revolves around deciding whether to paint a wall now or skip it, recalculating the total cost accordingly.

The key steps executed in the provided Java code are:

1. Initialize currentCost and nextCost arrays, ensuring that nextCost starts with a base state where no painting cost is incurred.
2. Process each wall in reverse order using a nested loop, where the inner loop calculates the cost for all possible remaining walls.
3. Within the nested loop, determine the cost to paint the current wall, adding to the cost calculated from a future state (nextCost), which is modified by the duration the current paint will last.
4. Compare the cost of painting the current wall to the cost of skipping it, updating the currentCost for the number of remaining walls.
5. After evaluating each wall, shift the costs from currentCost to nextCost to prepare for the next iteration.
6. At the completion of loops, return the cost to paint all walls stored in currentCost[count], which represents the final state where all walls are considered.

This approach ensures minimal cost computation through dynamic programming by iteratively refining the cost of painting each subset of walls based on decisions made for previous walls.

class Solution:
    def calculateMinCost(self, price: List[int], duration: List[int]) -> int:
        num_tasks = len(price)
        current_min = [0] * (num_tasks + 1)
        backup_min = [float('inf')] * (num_tasks + 1)
        backup_min[0] = 0
    
        for task in range(num_tasks - 1, -1, -1):
            current_min = [0] * (num_tasks + 1)
            for remaining in range(1, num_tasks + 1):
                do_task = price[task] + backup_min[max(0, remaining - 1 - duration[task])]
                skip_task = backup_min[remaining]
                current_min[remaining] = min(do_task, skip_task)
    
            backup_min = current_min
    
        return current_min[num_tasks]

This Python solution involves a dynamic programming approach to minimize the cost of painting walls given a list of prices and durations. The function calculateMinCost takes two parameters: price and duration, both lists of integers, representing the cost and the number of days required for each painting task respectively.

• Initialize two lists, current_min and backup_min, to track the minimal costs dynamically.
• Iterate backward through the tasks using a loop to decide, for each task, whether to execute or skip it.
• For each task, reset the current_min list to store new computed values.
• Within each task iteration, loop through a range from 1 to num_tasks + 1. Compute and compare two potential costs:
  • do_task: Perform the current task and add its cost to the minimized cost of subsequent tasks adjusted by its duration.
  • skip_task: Skip the current task, hence taking the minimized cost from the backup_min.
• Decide the minimum cost for current_min[remaining] by comparing the costs of do_task and skip_task.
• Assign current_min to backup_min at the end of each main task iteration to prepare for the next iteration.

Finally, the function returns current_min[num_tasks], which represents the minimum cost to complete all painting tasks. This solution achieves optimal painting strategy calculation using dynamic programming techniques to handle overlapping subproblems efficiently.